A good estimate would be 44. 17*1.61*1.61. In actual fact I have 55 parkruns within a 55 km radius, so slightly more spread out that the 17 mile radius which includes a couple of large cities.
Solution of the Week #437 - Rugby Scores part 2
To begin with, since we are only interested in the parity, we can calculate each subsequent value modulo 2, returning a 0 for an even number and a 1 for an odd number:
1,0,0,1,0,1,1,1,0, etc (let’s call this sequence b(n)).
Since each value only depends on the previous 6 values, we can return a 7 digit binary number for each value (using the fact that b of negative n is 0, to calculate the first few):
0000001, 0000010, 0000100, 0001001, 0010010, 0100101, etc
And then convert these into their decimal equivalents:
1,2,4,9,18,37,75,23,etc (let’s call this sequence c(n)).
Since whenever a particular number appears in this sequence it will be followed by the same number, and since there are only 128 possible values, from 0 to 127, using the pigeonhole principle, we know that the sequence must develop into a repeating loop, and in at most 128 steps.
We know that c(0)=1, and it doesn’t take much work on a spreadsheet or even by hand, to show that c(63)=1 also. So the sequence c(n), and therefore also the sequence b(n), repeats every 63 steps.
All that remains is to look at 63 consecutive values in b(n) and count how many are 0 and how many are 1.
31 are 1 and 32 are 0, and so the proportion of numbers in the a(n) Rugby sequence that are odd is exactly 31/63, or about 49.2%.
Solution of the Week #436 - Rugby Scores
There are a couple of ways to approach this. The most obvious way is to find all of the combinations of 3,5,7 can add to 23, then work out how many permutations there are of each. There are only 4 combinations: (7,7,3,3,3), (7,5,5,3,3), (5,5,5,5,3), (5,3,3,3,3,3,3).
The first combination has 5 numbers, split into 2 of one and 3 of another, so the number of permutations is 5!/(2!3!), which is 10. Doing the same for the other combinations, you get 30, 5 and 7 permutations respectively, so 52 ways in total.
A nice clever spreadsheet shortcut that I discovered is to let a(0)=1 and for anything else a(n)=a(n-7)+a(n-5)+a(n-3).
This recursive relationship very quickly reveals that 1,2 and 4 are impossible, that a(8)=2, that a(23)=52.
To see why this works, consider the score of 23. The scores that make it up must end with either 3, 5, or 7, and so before that final score is added, you would have a score of either 16, 18 or 20. And so the number of ways of making 23 must be the sum of the number of ways of making 16, 18 or 20.
As an interesting footnote, the smallest n for which a(n) requires a further digit follows an interesting pattern:
a(16)=10
a(26)=102
a(36)=1045
a(46)=10806
a(56)=112131
a(66)=1164641
a(76)=12098938
a(86)=125695132
But that’s where the pattern ends, as a(95) gives a ten-digit number of ways.
Solution of the Week #435 - A Satisfying Cancellation
The complicated expression turns out to have the value of precisely 3.
Call the sum of the two square roots ‘S’, and the difference ‘D’, the expression becomes S*sqrt(D/S)
Although S*D doesn’t appear anywhere in the expression, we will evaluate it in case it becomes useful later.
(sqrt(11)+sqrt(2))* (sqrt(11)-sqrt(2)) = 11+sqrt(22)–sqrt(22)-2 = 9.
If we look at the fraction under the square root, we can multiply top and bottom by D without changing its value. The top becomes D^2 and the bottom becomes 9. Then the square root (D^2/9) means that the whole square root expression can be replaced with D/3. Then the entire expression is just S*D/3, which of course is just 9/3 = 3.
Alternatively, change the left hand term S into the square root of S^2. Then you can combine the two square roots so that the overall expression is sqrt(S^2*D/S) or sqrt(S*D). As we saw above S*D is just 9, so sqrt(S*D)=3.
Solution of the Week #434 - Four Guesses
Your first guess is 2. Assuming you’re not correct, I had 1, 3 or 4. I must change to 2, 3 or 4.
Your second guess is 3. If your guess was incorrect, I was at either 2 or 4 and so must change to 1 or 3.
You guess 3 again. You either guessed correctly or I was at 1. I must now change to 2.
You guess 2. Checkmate!
By symmetry your guesses might have been 3,2,2,3 instead.
Solution of the Week #433 - Scores
The number denotes the alphabetical position of the first letter in alphabetical order that appears in the team name. For example Leicester City doesn’t contain a or b but does contain a c, hence their score is 3.
Therefore Everton beat Arsenal 5-1.
I’d love tickets to Luton Town versus Portsmouth!
Solution of the Week #432 - Ace of Base
The highest value is 4111111 in base 5, which is worth 66406 in decimal.
Solution of the Week #431 - Simplify
sqrt(sqrt(80)+sqrt(81))
= sqrt(4*sqrt(5) + 9)
= sqrt(4 + 4*sqrt(5) + 5)
= sqrt(2^2 + 2*2*sqrt(5) + sqrt(5)^2)
= sqrt(2 + sqrt(5))^2
= 2 + sqrt(5)
Solution of the Week #430 - 1x1x1 Rubik's Cube
The perhaps surprising answer is that every scramble can be solved in at most TWO adjacent colour swaps. The way I approached it was to separate the scrambles into three possible cases, based on the number of opposite pairs present in the scramble. This can be 0, 1 or 3. There can’t be exactly two opposite pairs present because if there were, the third opposite pair would also be there!
Of the 30 possible distinct cube arrangements, 16 are of the ‘no opposite pairs’ type, 12 have exactly one opposite pair and 2 have all three opposite pairs.
When there are three opposite pairs, you either already have the solved state or a complete mirror image of the solved state. For the mirror image you can just swap the blue and white, and the green and yellow, for instance, and you have solved the cube.
For the case where there is one opposite pair, let’s assume without loss of generality that the red and orange are opposite, but that green-blue and white-yellow are not. Looking at the RWB vertex, either it is already clockwise, in which case swapping GY will solve the cube, or RWB is anti-clockwise, in which case swapping WB will solve the cube. If Blue and White are opposite, then the RWB vertex won’t exist, but the RBY vertex will, and the same situation arises. So for the 1 opposite pair case, one swap is always enough.
For the case where none of the opposite pairs start off opposite, if you swap any one side so that it IS opposite its partner, for instance by swapping the Orange to the position opposite the Red, and you will reduce to the case where there is exactly one opposite pair present (the RO pair), and as we have already seen, that case can always be solved in just one move.
QED.
Solution of the Week #429 - Strange Stamps
There are three ways to do it:
1, 2, 5, 8, 11, 12, 13
1, 3, 4, 9, 10, 12, 13
1, 3, 5, 7, 8, 17, 18
Solution of the Week #428 - Doubly Generated Hexagon
Solution of the Week #427 - House Numbers part 2
There are 68 houses on the street.
Oddy lives at number 31 and Evelyn lives at number 56.
The odd numbers from 1 to 29 add to 225.
The odd numbers from 33 to 67 add to 900.
The even numbers from 2 to 54 add to 756.
The even numbers from 58 to 68 add to 378.
Solution of the Week #426 - House Numbers
The sum of the first n odd numbers is simply n^2. If I live at the kth house (ie house number 2k-1) then the houses to the left of it add to (k-1)^2. For example if I lived at number 7 (the fourth odd number) 1+3+5 is equal to 3^2 = 9.
For the houses to the right, if there are n houses altogether then we will have n^2 – k^2. This needs to be equal to (k-1)^2.
Effectively then, we are looking for two consecutive square numbers (k-1)^2 and k^2, which sum to another square number n^2.
Solutions exist when k is 1, 4, 21, 120, 697 etc. Only one of these corresponds to a three digit house number. The 120th odd number is 239. Therefore I live at number 239 in a street with houses up to number 337. House numbers either side of me add up to 14161.
Solution of the Week #425 - Trisected
Consider a simplified generalised version of the figure as below, scaling the figure down by a factor of 15 and letting the 33 be an unknown value for the time being.
From the angle bisector theorem, AB/BD = AF/FD. I’ve called the length of BD ‘a’ hence AB = ax.
ABD is a right-angled triangle, so by Pythagoras we have a^2+(ax)^2=(x+1)^2
a^2(x^2+1)=(x+1)^2
a^2=((x+1)^2)/(x^2+1)
We’ll also use the fact that triangle ABD is similar to triangle BDE, since they share the angle at D and they both have a right angle.
h/a=ax/(x+1)
h=xa^2/(x+1)
But we have another expression for a^2 above, so let’s use that:
h=x(x+1)/(x^2+1)
Now, here’s the clever step. We could have gone through the exact same steps from the right-hand side instead, and found a value for h in terms of y, and we would have ended up with the exact same expression. So it’s also true that:
h=y(y+1)/(y^2+1)
Now, returning to the specific puzzle, we’ll keep the DF=1, but now allow x=33/15, so we just have a scaled down version of the real puzzle.
Plugging x=33/15 into the expression gives us h=88/73.
88/73=y(y+1)/(y^2+1)
Cross-multiplying and multiplying out we get the quadratic:
15y^2-73y+88=0
This has the two solutions 33/15 and 40/15. We already knew about 33/15 as that was the value of x that we plugged in. So y=40/15.
Rescaling so that FD is once again equal to 15, we find that the missing value from the puzzle is 40.
There is arithmetically an easier method but it’s harder to see how it’s justified or why it works. And that is that the locus of points that are a constant ratio from C and F respectively is a circle whose diameter is AD. Once you know that, calculating the missing length is relatively straightforward.
?/15 = (?+48)/33
33? = 15? = 720
18? = 720
? = 40.
Solution of the Week #424 - Sock Drawer
The ‘G’ pair of socks are touching.
Solution of the Week #423 - Triangle 29
If the triangle has base 5 and height 2, then it will have hypotenuse of square root of 29. The smaller copies will therefore have hypotenuse of 1 and can fit together as below:
Solution of the Week #422 - Boggle E
Badge, bead, beck, café, code, complied, deaf, deck, face, faced, fade, gecko, head, heck, lied, life, mode.
Solution of the Week #421 - Odd Prime Balance
The lowest solution is achieved when T is 6 and U is 24. To find the values of V to Z, just choose the lowest number for which p, p+6 and p+24 are all prime, then the next lowest without overlapping, etc. So, our V to Z are 5, 7, 17, 37 and 47.
Our A to O will then be:
A = 5
B = 7+6 = 13
C = 17+24 = 41
D = 37
E = 47+6 = 53
F = 5+24 = 29
G = 7
H = 17+6 = 23
I = 37+24 = 61
J = 47
K = 5+6 = 11
L = 7+24 = 31
M = 17
N = 37+6 = 43
O = 47+24 = 71
Giving a total of 489, using 15 of the first 20 prime numbers.
Thanks to Graham Holmes for finding this optimal solution.
Solution of the Week #420 - Fourth Powers and Fourth Fractions
The initial step is to try to find a useful factorisation of the general form (n^4 + 1/4).
We will aim to factorise into two factors as below:
(n^2 +mn +1/2)(n^2 –mn +1/2)
Clearly the first terms in each bracket will combine to give the sought after n^4, and the last in each bracket will give the 1/4 we need. The fact that we have +mn and -mn means that the n^3 terms will cancel, as will the n^1 terms. We must then choose a value of m such that the n^2 terms disappear too.
Just showing the n^2 terms, we have (1/2)n^2 +(1/2)n^2 =(m^2)n^2.
m^2 = 1. So m is +/-1 (but since we have +m in one bracket and -m in the other it makes no difference which we choose).
So (n^4 + 1/4) = (n^2 +n +1/2)(n^2 -n +1/2)
We still need some further manipulation before it becomes useful.
If we take the first bracket and rewrite it as:
(n^2 +n +1/4 +1/4)
Then the first three terms form a perfect square:
((n+(1/2))^2 +1/4)
Doing the same for the second bracket we get
((n-(1/2))^2 +1/4)
To see how this is useful, let’s rewrite some of the terms of the initial question in this new form:
(1^4 +1/4) = ((1.5)^2 +1/4)((0.5)^2 +1/4)
(2^4 +1/4) = ((2.5)^2 +1/4)((1.5)^2 +1/4)
(3^4 +1/4) = ((3.5)^2 +1/4)((2.5)^2 +1/4)
The term ((1.5)^2 +1/4) will appear once in the numerator and once in the denominator, and so will cancel out. This will be true for almost all of the terms in the expression and we will just be left with: ((10.5)^2 +1/4)/((0.5)^2 +1/4)
To aid mental calculation, we can rewrite these two terms in the previous form:
(10^2 +10 +1/2)/(0^2 +0 +1/2)
110.5 *2
And so the solution is precisely 221.
Solution of the Week #419 - Minimum Triangle
It is tempting to think that the triangle would need to be Pythagorean. Certainly for any Pythagorean triangle, the values of a and b would be whole numbers, and c would be a rational number. Suitably scaled up, this too can become a whole number. For instance, if we started with a 3-4-5 triangle, a would be 2 and b would be 3. c would work out to be 2.4. Scaling everything by 5 would give a solution of (10,15,12) with a total of 37. This is not however the minimum solution.
It's a nice little fact that the area of such a triangle is simply a*b. It’s simple enough to prove with a bit of algebra; I’ll leave it to the reader to try it. And if you find a nice geometric proof, let me know as thus far one has eluded me!
That being the case, (a+b)*c/2, the standard way of finding the area, must be equal to a*b. Rearranging, we need to find two numbers a and b such that their product divided by their sum is either a whole number or a half number. The first such pair is 2 and 6, yielding the minimum solution of (2,6,3). The area is 12, which is both 2*6 and (2+6)*3/2. So the answer is 11. The resulting triangle is certainly not Pythagorean, having sides (2*sqrt(7)-2, 2*sqrt(7)+2,8).